Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.

First some notational specifications: Let us define $\tau$ as the topology of $X$ $\tau_{Y}=\{U \cap Y; U \in \tau \}$; $\tau_{A}^Y=\{V \cap A; V \in \tau_{Y} \}$ and $\tau_{A}^X=\{U \cap A; U \in \tau \}$

Let us show that $\tau_{A}^Y=\tau_{A}^X$

$\subset$: Let $V \cap A$ be an element of $\tau_{A}^Y$ such that $V \in \tau_{Y}$

Then, there exists $U\in \tau$ such that $V=U\cap Y$

$V\cap A=(U\cap Y)\cap A=U\cap (Y\cap A)= U\cap A$

Hence: $V\cap A \in \tau_{A}^X$ and so: $\tau_{A}^Y \subset \tau_{A}^X$

$\supset$: Let $U \cap A \in \tau_{A}^X$

$U\in \tau \iff U \text{ open in } X$

But $U \cap Y$ open in $Y \iff U \cap Y \in \tau_{Y}$

Hence $(U \cap Y) \cap A \in \tau_{A}^Y$ since $U \cap Y \in \tau_{Y}$

Therefore $U \cap A \in \tau_{A}^Y$

Hence: $\tau_{A}^X \subset \tau_{A}^Y$

Therefore: $\tau_{A}^Y=\tau_{A}^X$

  • $\begingroup$ When you conclude in the third last line $U \cap A \in \tau ^Y_A$. Just mention before that $(U \cap Y) \cap A =U \cap A$. Otherwise the proof is perfect. $\endgroup$ – Babai Dec 30 '13 at 13:07
  • $\begingroup$ @Susobhan Ok. Thank you $\endgroup$ – user43418 Dec 30 '13 at 13:10

Let $\tau_{X}$ and $\tau_{Y}$ be the topologies of A as subspace of X and Y.

If $U \in \tau_{X} \Rightarrow \exists B $ open in X such that $U = B \cap A = (B\cap Y) \cap A$ and then $U \in \tau_{Y}$.

If $U \in \tau_{Y} \Rightarrow \exists B $ open in Y such that $U = B \cap A \Rightarrow \exists C $ open in X such that $ B = C \cap Y \Rightarrow U = (C \cap Y)\cap A = C \cap ( Y \cap A) = C \cap A.$

Then $\tau_{X} =\tau_{Y}$.


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